Problem: Given a string of length n s = s1, s2,…, sn and m queries. Each query qk (1 ≤ k ≤ m) is one of four types, "L ++", "L-", "R ++", "R-", and l [for the kth query qk. k] and r [k] are defined below.

* L ++: l [k] = l [k-1] + 1, r [k] = r [k-1]

* L-: l [k] = l [k-1] -1, r [k] = r [k-1]

* R ++: l [k] = l [k-1], r [k] = r [k-1] +1

* R-: l [k] = l [k-1], r [k] = r [k-1] -1




However, l [0] = r [0] = 1.

At this time, how many kinds of character strings are created for m substrings sl [k], sl [k] + 1,…, sr [k] -1, sr [k] (1 ≤ k ≤ m). Answer if you can.

Constraints

* The string s consists of a lowercase alphabet

* 1 ≤ n ≤ 3 x 105

* 1 ≤ m ≤ 3 × 105

* qk (1 ≤ k ≤ m) is one of "L ++", "L-", "R ++", "R-"

* 1 ≤ l [k] ≤ r [k] ≤ n (1 ≤ k ≤ m)

Input

Input is given in the following format

> n m
> s
> q1
> q2
>…
> qm
>

Output

Output the solution of the problem on one line

Examples

Input

5 4
abcde
R++
R++
L++
L--


Output

3


Input

4 6
abab
R++
L++
R++
L++
R++
L++


Output

4


Input

10 13
aacacbabac
R++
R++
L++
R++
R++
L++
L++
R++
R++
L--
L--
R--
R--


Output

11

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define all(v) begin(v), end(v)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int const MOD = 1000000007;

namespace std {
    bool operator<(const P& a, const P& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

// Rolling Hash
struct RollingHash {
public:
    long long base_1, base_2, mod_1, mod_2;
    vector<long long> hash1, hash2, pow1, pow2;
    RollingHash(string s) : base_1(1013), base_2(1007), 
        mod_1(1000000007), mod_2(1000000009),
        hash1(s.length()+1,0), hash2(s.length()+1,0),
        pow1(s.length()+1,1), pow2(s.length()+1,1) {
            for(int i=0; i<s.length(); i++) {
                hash1[i+1] = (hash1[i]+s[i]) * base_1 % mod_1;
                hash2[i+1] = (hash2[i]+s[i]) * base_2 % mod_2;
                pow1[i+1] = pow1[i] * base_1 % mod_1;
                pow2[i+1] = pow2[i] * base_2 % mod_2;
            }
        }

    // calc str[l,r) hash
    pair<long long, long long> get(int l, int r) {
        long long t1 = ((hash1[r] - hash1[l] * pow1[r-l]) % mod_1 + mod_1) % mod_1;
        long long t2 = ((hash2[r] - hash2[l] * pow2[r-l]) % mod_2 + mod_2) % mod_2;
        return make_pair(t1, t2);
    }
};

signed main() {
    int n, m; cin >> n >> m;
    string s, inp; cin >> s;

    RollingHash RH(s);
    set<pii> strset;
    int l = 0, r = 0;
    rep(i,0,m) {
        cin >> inp;
        if(inp == "R++") r++;
        if(inp == "R--") r--;
        if(inp == "L++") l++;
        if(inp == "L--") l--;
        pii p = RH.get(l,r+1);
        // printf("l = %lld, r = %lld, get = (%lld, %lld)\n", l, r, p.fr, p.sc);
        strset.insert(p);
    }
    cout << strset.size() << endl;
    return 0;
}